How do I graph the hyperbola with the equation #4x^2−y^2+4y−20=0?#?

1 Answer
Aug 14, 2018

Answer:

Please see the explanation below

Explanation:

You need the center, the vertices and the asymptotes.

The equation is

#4x^2-y^2+4y-20=0#

Completing the square

#4x^2-(y^2-4y+4)=20-4=16#

Dividing by #16#

#4/16x^2-(y-2)^2/16=1#

#(x)^2/4-(y-2)^2/16=1#

Comparing this to the general equation of a hyperbola

#(x-h)^2/a^2-(y-k)^2/b^2=1#

The center is #C=(h,k)=(0,2)#

The vertices are

#A=(h+a,k)=(0+2,2)=(2,2)#

and

#A'=(h-a,k)=(0-2,2)=(-2,2)#

And the asymptotes are

#(y-2)^2/16=(x)^2/4#

#y-2=+-(2x)#

#y-2=2x# and #y-2=-2x#

Now, you can plot the graph

graph{(4x^2-y^2+4y-20)(y-2-2x)(y-2+2x)=0 [-15.77, 16.27, -5.8, 10.2]}